TITLE: Proof of Vandermonde's Identity using a "different approach" using complex integration
QUESTION [5 upvotes]: Hi I'd like to know if the following proof of Vandermonde's Identity is correct (is really easy):
Let $m,n,r$ be natural numbers such that $r\le \min \{m,n\}$.  The Vandermonde's Identity gives us that $$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$
$\bf{Solution}$ Let $\gamma$ be the unit circle $\lvert z\lvert =1$. Now since $${m \choose r}= \frac{1}{2\pi i}\int _{\gamma}\frac{(1+z)^m}{z^{r+1}} dz$$
Then 
\begin{align}  2\pi i \binom{m+n}r= \int _{\gamma}\frac{(1+z)^{m+n}}{z^{r+1}} dz &= \int _{\gamma}\frac{(1+z)^{n}}{z^{r+1}}\sum_{k=0}^m  {m \choose k}z^k \,dz\\
&= \sum_{k=0}^m  { m\choose k}\int_\gamma \frac{(1+z)^n}{z^{r-k+1}} dz \\
&= \sum_{k=0}^m 2\pi i { m\choose k} {n\choose r-k} \end{align}
For all $k>r$, we set ${n\choose r-k}=0$, i.e., the left-hand side is zero and imply the desired identity
$$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$
Thanks :)

REPLY [3 votes]: To keep the question off the unanswered list. Yes, your approach is correct.
See Complex Analysis proof of multinomial expression for something in the same spirit.